# How do you verify the identity: [sin(x) / csc(x) - 1 ] = [ sin (x) + 1 / cot^2 (x) ]?

Apr 27, 2015

Short answer: there is no identity. Here is the logic.

Firstly, it's always advisable to determine the domain of the functions on both sides of the equation.

On the left the restriction is $\csc \left(x\right)$ exists and $\csc \left(x\right) \ne 0$. Since $\csc \left(x\right) = \frac{1}{\sin} \left(x\right)$ (by definition), existence of $\csc \left(x\right)$ means $\sin \left(x\right) \ne 0$ or $x \ne \pi \cdot N$. Also notice that $\csc \left(x\right)$ never equal to zero.

On the right we need: $\cot \left(x\right)$ should exist and $\cot \left(x\right) \ne 0$.
Since $\cot \left(x\right) = \cos \frac{x}{\sin} \left(x\right)$ (by definition), its existence means $\sin \left(x\right) \ne 0$, that is $x \ne \pi \cdot N$, and not being equal to zero means $\cos \left(x\right) \ne 0$, that is $x \ne \frac{\pi}{2} + \pi \cdot N$.

Before going any further, it makes sense to check if the identity is true for some simple case. Let's say, $x = \frac{\pi}{4}$ (that is, ${45}^{0}$).
In this case
$\sin \left(x\right) = \cos \left(x\right) = \frac{\sqrt{2}}{2}$;
$\tan \left(x\right) = \cot \left(x\right) = 1$.
Therefore, the left side evaluates to
$\left(\frac{\sqrt{2}}{2}\right) : \left(\frac{2}{\sqrt{2}}\right) - 1 = - \frac{1}{2}$
The right side evaluates to
$\frac{\sqrt{2}}{2} + \frac{1}{1} ^ 2 = \frac{\sqrt{2}}{2} + 1$

Obviously, these two expressions are not equal to each other for this particular value of $x = \frac{\pi}{4}$. So, there is no general equality between them.